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Find the sum of the x values which solve the system of equations \(3y - 49 = 0 \\ x^2 + y^2 - 281 = 0\)

 
waffles  Nov 14, 2017

Best Answer 

 #1
avatar+1961 
+3

3y - 49 = 0

x^2 + y^2 - 281 = 0

 

First I would isolate the variable (y) in the first equation

 

3y - 49 = 0

Add 49 on both sides

3y = 49

Divide by 3 on both sides

y = 49/3

 

Then take your new equation and plug it into the second equation

 

x^2 + (49/3)^2 - 281 = 0

Simplify

x^2 + (2401/9) - 281 = 0

Subtract 281 from the fraction

x^2 + (-128/9) = 0

Add (128/9) on both sides

x^2 = (128/9)

Root on both sides

|x| = \(\sqrt{\frac{128}{9}}\)

Simplify (the x is a multiplication)

|x| = \(\frac{\sqrt{64 x 2}}{3}\)

Simplify again

|x| = \(\frac{8\sqrt{2}}{3}\)

Then the absolute value turns the other side into ±

x = ±\(\frac{8\sqrt{2}}{3}\)

And there's your answer! :) Hope this helps

 
saseflower  Nov 14, 2017
edited by saseflower  Nov 14, 2017
Sort: 

4+0 Answers

 #1
avatar+1961 
+3
Best Answer

3y - 49 = 0

x^2 + y^2 - 281 = 0

 

First I would isolate the variable (y) in the first equation

 

3y - 49 = 0

Add 49 on both sides

3y = 49

Divide by 3 on both sides

y = 49/3

 

Then take your new equation and plug it into the second equation

 

x^2 + (49/3)^2 - 281 = 0

Simplify

x^2 + (2401/9) - 281 = 0

Subtract 281 from the fraction

x^2 + (-128/9) = 0

Add (128/9) on both sides

x^2 = (128/9)

Root on both sides

|x| = \(\sqrt{\frac{128}{9}}\)

Simplify (the x is a multiplication)

|x| = \(\frac{\sqrt{64 x 2}}{3}\)

Simplify again

|x| = \(\frac{8\sqrt{2}}{3}\)

Then the absolute value turns the other side into ±

x = ±\(\frac{8\sqrt{2}}{3}\)

And there's your answer! :) Hope this helps

 
saseflower  Nov 14, 2017
edited by saseflower  Nov 14, 2017
 #2
avatar
0

You made a small mistake:

x = +or-​ 8sqrt(2)/3

 
Guest Nov 14, 2017
 #3
avatar+1961 
+3

Thank you so much! I fixed it

 
saseflower  Nov 14, 2017
 #4
avatar+78753 
+2

And..as saseflower would say....the sum of the x values that solve this system is just  0

 

 

cool cool cool

 
CPhill  Nov 14, 2017

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